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Question Number 14633 by Tinkutara last updated on 03/Jun/17

Solve the equation: (1 − tanθ)(1 + sin2θ) = 1 + tanθ

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\left(\mathrm{1}\:−\:\mathrm{tan}\theta\right)\left(\mathrm{1}\:+\:\mathrm{sin2}\theta\right)\:=\:\mathrm{1}\:+\:\mathrm{tan}\theta \\ $$

Commented by myintkhaing last updated on 03/Jun/17

((1−tanθ)/(1+tanθ)) (1+sin2θ) = 1 ((cos2θ)/(1+sin2θ)) (1+sin2θ) = 1

$$\frac{\mathrm{1}−{tan}\theta}{\mathrm{1}+{tan}\theta}\:\left(\mathrm{1}+{sin}\mathrm{2}\theta\right)\:=\:\mathrm{1} \\ $$$$\frac{{cos}\mathrm{2}\theta}{\mathrm{1}+{sin}\mathrm{2}\theta}\:\left(\mathrm{1}+{sin}\mathrm{2}\theta\right)\:=\:\mathrm{1} \\ $$

Answered by ajfour last updated on 03/Jun/17

(1−tan θ)(1+((2tan θ)/(1+tan^2 θ)))=1+tan θ since tan θ≠i , we multiply by 1+tan^2 θ, further let tan θ=x; then (1−x)(1+x)^2 =(1+x)(1+x^2 ) ⇒ (1+x)(1−x^2 )=(1+x)(1+x^2 ) or (1+x)(1+x^2 −1+x^2 )=0 ⇒ x^2 (1+x)=0 this means x=tan θ=0,−1 ⇒ 𝛉=n𝛑 or 𝛉=n𝛑−(𝛑/4) .

$$\left(\mathrm{1}−\mathrm{tan}\:\theta\right)\left(\mathrm{1}+\frac{\mathrm{2tan}\:\theta}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \theta}\right)=\mathrm{1}+\mathrm{tan}\:\theta \\ $$$${since}\:\mathrm{tan}\:\theta\neq{i}\:\:,\:{we}\:{multiply}\:{by} \\ $$$$\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \theta,\:\:{further}\:{let}\:\mathrm{tan}\:\theta=\boldsymbol{{x}}; \\ $$$${then} \\ $$$$\:\:\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)^{\mathrm{2}} =\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:\:\left(\mathrm{1}+{x}\right)\left(\mathrm{1}−{x}^{\mathrm{2}} \right)=\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right) \\ $$$${or}\:\:\:\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{1}+{x}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow\:\:{x}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)=\mathrm{0} \\ $$$${this}\:{means} \\ $$$$\:\:{x}=\mathrm{tan}\:\theta=\mathrm{0},−\mathrm{1} \\ $$$$\Rightarrow\:\:\:\:\boldsymbol{\theta}=\boldsymbol{{n}\pi}\:{or}\:\boldsymbol{\theta}=\boldsymbol{{n}\pi}−\frac{\boldsymbol{\pi}}{\mathrm{4}}\:\:. \\ $$$$ \\ $$

Commented by ajfour last updated on 03/Jun/17

the way the question is put, we need not substitute tan θ=((sin θ)/(cos θ)) , or 𝛉=nπ−(π/4) is lost .

$${the}\:{way}\:{the}\:{question}\:{is}\:{put}, \\ $$$${we}\:{need}\:{not}\:{substitute}\: \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}\:,\:{or}\:\boldsymbol{\theta}={n}\pi−\frac{\pi}{\mathrm{4}}\:{is}\:{lost}\:. \\ $$

Commented by Tinkutara last updated on 03/Jun/17

Thanks Sir for the exact solution!

$$\mathrm{Thanks}\:\mathrm{Sir}\:\mathrm{for}\:\mathrm{the}\:\mathrm{exact}\:\mathrm{solution}! \\ $$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 03/Jun/17

1+sin2θ=sin^2 θ+cos^2 θ+2sinθcosθ= =(sinθ+cosθ)^2 ((1−tgθ)/(1+tgθ))=((cosθ−sinθ)/(cosθ+sinθ)) ⇒((cosθ−sinθ)/(cosθ+sinθ)).(sinθ+cosθ)^2 =1 1)cosθ+sinθ≠0⇒tgθ≠−1⇒θ≠kπ+((3π)/4) 2)cos^2 θ−sin^2 θ=1⇒cos2θ=1⇒2θ=2mπ ⇒θ=mπ . m∈Z.

$$\mathrm{1}+{sin}\mathrm{2}\theta={sin}^{\mathrm{2}} \theta+{cos}^{\mathrm{2}} \theta+\mathrm{2}{sin}\theta{cos}\theta= \\ $$$$=\left({sin}\theta+{cos}\theta\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}−{tg}\theta}{\mathrm{1}+{tg}\theta}=\frac{{cos}\theta−{sin}\theta}{{cos}\theta+{sin}\theta} \\ $$$$\Rightarrow\frac{{cos}\theta−{sin}\theta}{{cos}\theta+{sin}\theta}.\left({sin}\theta+{cos}\theta\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\left.\mathrm{1}\right){cos}\theta+{sin}\theta\neq\mathrm{0}\Rightarrow{tg}\theta\neq−\mathrm{1}\Rightarrow\theta\neq{k}\pi+\frac{\mathrm{3}\pi}{\mathrm{4}} \\ $$$$\left.\mathrm{2}\right){cos}^{\mathrm{2}} \theta−{sin}^{\mathrm{2}} \theta=\mathrm{1}\Rightarrow{cos}\mathrm{2}\theta=\mathrm{1}\Rightarrow\mathrm{2}\theta=\mathrm{2}{m}\pi \\ $$$$\Rightarrow\theta={m}\pi\:.\:\:\:{m}\in{Z}. \\ $$

Commented by Tinkutara last updated on 03/Jun/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

Answered by 1kanika# last updated on 03/Jun/17

Commented by 1kanika# last updated on 03/Jun/17

sorry..... θ=nπ where n belongs to Z

$$\mathrm{sorry}.....\: \\ $$$$\:\theta=\mathrm{n}\pi\:\:\: \\ $$$${where}\:\:{n}\:\mathrm{belongs}\:\mathrm{to}\:\:\mathrm{Z} \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 03/Jun/17

my answer is: θ=mπ,and yours:θ=nπ there is no diffrence between this answers.

$${my}\:{answer}\:{is}:\:\theta={m}\pi,{and}\:{yours}:\theta={n}\pi \\ $$$${there}\:{is}\:{no}\:{diffrence}\:{between}\:{this}\: \\ $$$${answers}. \\ $$

Commented by Tinkutara last updated on 03/Jun/17

Thanks but you lost one solution!

$$\mathrm{Thanks}\:\mathrm{but}\:\mathrm{you}\:\mathrm{lost}\:\mathrm{one}\:\mathrm{solution}! \\ $$

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